Solve 7+3/8-9i

$$\frac { 7 + 3 } { 8 - 9 i }$$

Evaluate

$$\frac{16}{29}+\frac{18}{29}i\approx 0.551724138+0.620689655i$$

Solution Steps

Add 7 and 3 to get 10.

$$\frac{10}{8-9i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 8+9i.

$$\frac{10\left(8+9i\right)}{\left(8-9i\right)\left(8+9i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{10\left(8+9i\right)}{8^{2}-9^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{10\left(8+9i\right)}{145}$$

Multiply 10 times 8+9i.

$$\frac{10\times 8+10\times \left(9i\right)}{145}$$

Do the multiplications in 10\times 8+10\times \left(9i\right).

$$\frac{80+90i}{145}$$

Divide 80+90i by 145 to get \frac{16}{29}+\frac{18}{29}i.

$$\frac{16}{29}+\frac{18}{29}i$$

Real Part

$$\frac{16}{29} = 0.5517241379310345$$

Solution Steps

Add 7 and 3 to get 10.

$$Re(\frac{10}{8-9i})$$

Multiply both numerator and denominator of \frac{10}{8-9i} by the complex conjugate of the denominator, 8+9i.

$$Re(\frac{10\left(8+9i\right)}{\left(8-9i\right)\left(8+9i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{10\left(8+9i\right)}{8^{2}-9^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{10\left(8+9i\right)}{145})$$

Multiply 10 times 8+9i.

$$Re(\frac{10\times 8+10\times \left(9i\right)}{145})$$

Do the multiplications in 10\times 8+10\times \left(9i\right).

$$Re(\frac{80+90i}{145})$$

Divide 80+90i by 145 to get \frac{16}{29}+\frac{18}{29}i.

$$Re(\frac{16}{29}+\frac{18}{29}i)$$

The real part of \frac{16}{29}+\frac{18}{29}i is \frac{16}{29}.

$$\frac{16}{29}$$